1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Quantum mechanics - spin operator eigenvalue probability?

  1. May 23, 2010 #1
    Hi,
    I have this problem on a past exam paper I am having some trouble with:

    "in the conventional basis of the eigenstates of the Sz operator, the spin state of a spin-1/2 particle is described by the vector:

    [tex] u = \left( \stackrel{cos a}{e^i^b sina} \right)[/tex] where a and B are constants.

    find the probability that a measurement of the y-component of the spin of the particle will yield the result [tex] 0.5\hbar [/tex] ."

    For the life of me I cannot work out how to write out matrices legibly on this thing, so I will summarize what is bothering me. I am given pauli matrices [tex] \sigma_x_,_y_,_z [/tex]that I cannot write out properly, and the spin operator is given by [tex] S_i = i\hbar\sigma_i [/tex].

    In the question I am given the vector u, which is apparently expressed in the basis of Sz eigenstates.

    Am I justified in putting this vector u into an eigenvalue equation [tex] S_{y} u = a_{y} u [/tex] ,

    where ay is my eigenvalue, when the vector I would be operating on is made from a basis of eigenstates of another operator (Sz)?

    I tried this and got two equations for ay, neither of which gives [tex] a_y = 0.5\hbar [/tex].

    does this mean I can conclude that there is zero probability of finding the y-component of the spin being equal to [tex] 0.5\hbar [/tex] ?

    or do I somehow have to wangle it so that I get another vector (not u) that is in the Sy eigenstate basis?

    thanks.



    PS. sorry, this crazy thing will not let me change something 5 lines up where I should have said
    "Am I justified in putting this vector u into an eigenvalue equation S_y u = 0.5\hbar u"
     
    Last edited: May 23, 2010
  2. jcsd
  3. May 23, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You tell me. If [itex]u[/itex] is an eigenvector of [itex]S_z[/itex], is it also an eigenvector of S_y[/itex]?
     
  4. May 23, 2010 #3
    are you actually asking me, do these two operators commute? and if their commutator [Sy, Sz] = 0, then yes they do?

    is that right, if the eigenvector of one operator is equal to the eigenvector of another, then the operators commute?
     
  5. May 23, 2010 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Yes. So, do [itex]S_z[/itex] and [itex]S_y[/itex] commute?
     
  6. May 23, 2010 #5
    not by my calculations, although I just woke up 5 minutes ago. I get [Sy,Sz] is equal to
    (hbar^2)/4 multiplied by a 2x2 matrix containing zeros on the leading diagonal and 2i elsewhere.
     
  7. May 23, 2010 #6

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Right, they don't commute; [itex][S_y,S_z]=i\hbar S_x[/itex]. So, [itex]u[/itex] is clearly not an eigenstate of [itex]S_y[/itex].

    What are the eigenstates of [itex]S_y[/itex]?
     
  8. May 23, 2010 #7
    i got that one of them was the column vector, say, x:
    x_1

    ix_1

    and the other vector y:
    y_1

    -iy_1

    but I was not convinced by this because, if the upper component is say, x1, and I choose x1 = 1, then if I go to normalize it i do
    1/sqrt[(1^2) + (i^2)] = 1/(1-1) = 0, so how can this be right if Im getting a normalization constant of zero?
     
  9. May 23, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You forgot to take the conjugate. You should be calculating [itex]\langle x|x \rangle = x^\dagger x[/itex], not [itex]x^T x[/itex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook